Orgo Question dat bootcamp

[JDHK]

Orgo is hard...

This attached pic. I tried drawing it, and picturing it in my head and cannot for the life of me admit that the two are the same..Lets call the carbon on the left side carbon #1. It has bromine attached to it which is going into the plane, and a methyl group on the plane. I visualized the rotation of the bromine group into the plane, and with that rotation, the methyl group rotates into the plane and the hydrogen is on the plane. All good so far. The problem is that when the whole molecule is rotated in that specific way, the other carbon #2 seems completely wrong. The methyl that was originally sticking out of the plane should now be on the plane (but it is still out of the plane), and the ethyl group should be out of the plane but it is into the plane, etc.

Does anyone get what I'm saying here? I'm not sure if I'm just absolutely wrong..

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[Illfavor]

Generally I assign R/S before making any decisions about stereochemistry.

 

[JDHK]

i'm not as good at orgo as you. lol i'm not quite sure if your response answers my question or if my question is just nonsense =(

 

[Illfavor]

i'm not as good at orgo as you. lol i'm not quite sure if your response answers my question or if my question is just nonsense =(

No problem. If you assign R/S you can convince yourself they are indeed the same. As far as the rotation goes: The bromine comes towards us into the plane and the methyl group (C1) therefore rotates backwards, with a dash, out of the plane. That's one half done. The Wedge methyl group on C3 doesn't really move, it's just that our orientation shifts so that C4/C5 now look like they're going away from us. Our viewpoint shifts up and left, so we are seeing "down" the C3 wedge.

 

[Ari Rezaei]

Orgo is hard...

This attached pic. I tried drawing it, and picturing it in my head and cannot for the life of me admit that the two are the same..Lets call the carbon on the left side carbon #1. It has bromine attached to it which is going into the plane, and a methyl group on the plane. I visualized the rotation of the bromine group into the plane, and with that rotation, the methyl group rotates into the plane and the hydrogen is on the plane. All good so far. The problem is that when the whole molecule is rotated in that specific way, the other carbon #2 seems completely wrong. The methyl that was originally sticking out of the plane should now be on the plane (but it is still out of the plane), and the ethyl group should be out of the plane but it is into the plane, etc.

Does anyone get what I'm saying here? I'm not sure if I'm just absolutely wrong..

You're right on the rotation for carbon #1, but this is a sigma bond. Both carbons are free to rotate as they please. You don't need to rotate the whole molecule, just the one carbon (or both separately if needed). Does this answer your question or am I missing it?

 

[JDHK]

Getting closer. Thanks so much for the response, and help, but I still am having trouble with one concept. (sry for the barrage of questions). So my question is if you have rotated our viewpoint on carbon 2, just as you have described, I was thinking we must rotate the entire molecule in the same way. Like it's just a static molecule that we are looking at it from a different perspective. Therefore the carbon 3 must rotate in the exact same way, which means since we're rotating the molecule out toward us, the ethyl group should now be wedged out at us, not into the plane as it shows.

Thanks SO much!!

 

[JDHK]

You're right on the rotation for carbon #1, but this is a sigma bond. Both carbons are free to rotate as they please. You don't need to rotate the whole molecule, just the one carbon. Does this answer your question or am I missing it?

Exactly answers my question. Thanks so much!!

 

[Ari Rezaei]

Exactly answers my question. Thanks so much!!

No problem